3.8 \(\int \frac {a+b \sec ^{-1}(c x)}{x} \, dx\)

Optimal. Leaf size=64 \[ \frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{2} i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right ) \]

[Out]

1/2*I*(a+b*arcsec(c*x))^2/b-(a+b*arcsec(c*x))*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+1/2*I*b*polylog(2,-(1/c/x+
I*(1-1/c^2/x^2)^(1/2))^2)

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Rubi [A]  time = 0.08, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5218, 4626, 3719, 2190, 2279, 2391} \[ \frac {1}{2} i b \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )+\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/x,x]

[Out]

((I/2)*(a + b*ArcSec[c*x])^2)/b - (a + b*ArcSec[c*x])*Log[1 + E^((2*I)*ArcSec[c*x])] + (I/2)*b*PolyLog[2, -E^(
(2*I)*ArcSec[c*x])]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4626

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> -Subst[Int[(a + b*x)^n/Cot[x], x], x, ArcCos[c
*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 5218

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCos[x/c])/x, x], x, 1/x] /; Fre
eQ[{a, b, c}, x]

Rubi steps

\begin {align*} \int \frac {a+b \sec ^{-1}(c x)}{x} \, dx &=-\operatorname {Subst}\left (\int \frac {a+b \cos ^{-1}\left (\frac {x}{c}\right )}{x} \, dx,x,\frac {1}{x}\right )\\ &=\operatorname {Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-2 i \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+b \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )-\frac {1}{2} (i b) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(c x)}\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac {1}{2} i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 59, normalized size = 0.92 \[ a \log (x)+\frac {1}{2} i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )+\frac {1}{2} i b \sec ^{-1}(c x)^2-b \sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSec[c*x])/x,x]

[Out]

(I/2)*b*ArcSec[c*x]^2 - b*ArcSec[c*x]*Log[1 + E^((2*I)*ArcSec[c*x])] + a*Log[x] + (I/2)*b*PolyLog[2, -E^((2*I)
*ArcSec[c*x])]

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arcsec}\left (c x\right ) + a}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x,x, algorithm="fricas")

[Out]

integral((b*arcsec(c*x) + a)/x, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x,x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(x)]Undef/Un
signed Inf encountered in limitLimit: Max order reached or unable to make series expansion Error: Bad Argument
Value

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maple [A]  time = 0.19, size = 86, normalized size = 1.34 \[ a \ln \left (c x \right )+\frac {i b \mathrm {arcsec}\left (c x \right )^{2}}{2}-b \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+\frac {i b \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsec(c*x))/x,x)

[Out]

a*ln(c*x)+1/2*I*b*arcsec(c*x)^2-b*arcsec(c*x)*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+1/2*I*b*polylog(2,-(1/c/x+
I*(1-1/c^2/x^2)^(1/2))^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -{\left (c^{2} \int \frac {\sqrt {c x + 1} \sqrt {c x - 1} \log \relax (x)}{c^{4} x^{3} - c^{2} x}\,{d x} - \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right ) \log \relax (x)\right )} b + a \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x,x, algorithm="maxima")

[Out]

-(c^2*integrate(sqrt(c*x + 1)*sqrt(c*x - 1)*log(x)/(c^4*x^3 - c^2*x), x) - arctan(sqrt(c*x + 1)*sqrt(c*x - 1))
*log(x))*b + a*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(1/(c*x)))/x,x)

[Out]

int((a + b*acos(1/(c*x)))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asec}{\left (c x \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asec(c*x))/x,x)

[Out]

Integral((a + b*asec(c*x))/x, x)

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